Detect cycle in an undirected graph using BFS

Given an undirected graph, how to check if there is a cycle in the graph? For example, the following graph has a cycle 1-0-2-1.
cycleGraph



We have discussed cycle detection for directed graph. We have also discussed a union-find algorithm for cycle detection in undirected graphs.. The time complexity of the union-find algorithm is O(ELogV). Like directed graphs, we can use DFS to detect cycle in an undirected graph in O(V+E) time. We have discussed DFS based solution for cycle detection in undirected graph.

In this article, BFS based solution is discussed. We do a BFS traversal of the given graph. For every visited vertex ‘v’, if there is an adjacent ‘u’ such that u is already visited and u is not parent of v, then there is a cycle in graph. If we don’t find such an adjacent for any vertex, we say that there is no cycle. The assumption of this approach is that there are no parallel edges between any two vertices.

We use a parent array to keep track of parent vertex for a vertex so that we do not consider visited parent as cycle.

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// C++ program to detect cycle in an undirected graph
// using BFS.
#include <bits/stdc++.h>
using namespace std;
  
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
  
bool isCyclicConntected(vector<int> adj[], int s,
                        int V, vector<bool>& visited)
{
    // Set parent vertex for every vertex as -1.
    vector<int> parent(V, -1);
  
    // Create a queue for BFS
    queue<int> q;
  
    // Mark the current node as visited and enqueue it
    visited[s] = true;
    q.push(s);
  
    while (!q.empty()) {
  
        // Dequeue a vertex from queue and print it
        int u = q.front();
        q.pop();
  
        // Get all adjacent vertices of the dequeued
        // vertex s. If a adjacent has not been visited,
        // then mark it visited and enqueue it. We also
        // mark parent so that parent is not considered
        // for cycle.
        for (auto v : adj[u]) {
            if (!visited[v]) {
                visited[v] = true;
                q.push(v);
                parent[v] = u;
            }
            else if (parent[u] != v)
                return true;
        }
    }
    return false;
}
  
bool isCyclicDisconntected(vector<int> adj[], int V)
{
    // Mark all the vertices as not visited
    vector<bool> visited(V, false);
  
    for (int i = 0; i < V; i++)
        if (!visited[i] && isCyclicConntected(adj, i,
                                         V, visited))
            return true;
    return false;
}
  
// Driver program to test methods of graph class
int main()
{
    int V = 4;
    vector<int> adj[V];
    addEdge(adj, 0, 1);
    addEdge(adj, 1, 2);
    addEdge(adj, 2, 0);
    addEdge(adj, 2, 3);
  
    if (isCyclicDisconntected(adj, V))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Output :

Yes

Time Complexity: The program does a simple BFS Traversal of graph and graph is represented using adjacency list. So the time complexity is O(V+E)



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